NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra Miscellaneous Exercise - Free Study Material ,CBSE Sample Papers, Books

NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra Miscellaneous Exercise

Page No 458:

Question 1:

Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.

Answer:

is a unit vector in the XY-plane, then

Here, θ is the angle made by the unit vector with the positive direction of the x-axis.

Therefore, for θ = 30°:

Hence, the required unit vector is

Question 2:

Find the scalar components and magnitude of the vector joining the points

Answer:

The vector joining the points

can be obtained by,

Hence, the scalar components and the magnitude of the vector joining the given points are respectively

and

Question 3:

A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.

Answer:

Let O and B be the initial and final positions of the girl respectively.

Then, the girl’s position can be shown as:

Now, we have:

By the triangle law of vector addition, we have:

Hence, the girl’s displacement from her initial point of departure is

Question 4:

, then is it true that

? Justify your answer.

Answer:

Now, by the triangle law of vector addition, we have

It is clearly known that

represent the sides of ΔABC.

Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.

Hence, it is not true that

Question 5:

Find the value of x for which

is a unit vector.

Answer:

is a unit vector if

Hence, the required value of x is

Question 6:

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

Answer:

We have,

Let

be the resultant of

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors

Question 7:

, find a unit vector parallel to the vector

Answer:

We have,

Hence, the unit vector along

Question 8:

Show that the points A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

The given points are A (1, –2, –8), B (5, 0, –2), and C (11, 3, 7).

Thus, the given points A, B, and C are collinear.

Now, let point B divide AC in the ratio

. Then, we have:

On equating the corresponding components, we get:

Hence, point B divides AC in the ratio

Question 9:

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are

externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Answer:

It is given that

It is given that point R divides a line segment joining two points P and Q externally in the ratio 1: 2. Then, on using the section formula, we get:

Therefore, the position vector of point R is

Position vector of the mid-point of RQ =

Hence, P is the mid-point of the line segment RQ.

Question 10:

The two adjacent sides of a parallelogram are

and

Find the unit vector parallel to its diagonal. Also, find its area.

Answer:

Adjacent sides of a parallelogram are given as:

and

Then, the diagonal of a parallelogram is given by

Thus, the unit vector parallel to the diagonal is

Area of parallelogram ABCD =

Hence, the area of the parallelogram is

square units.

Question 11:

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are

Answer:

Let a vector be equally inclined to axes OX, OY, and OZ at angle α.

Then, the direction cosines of the vector are cos α, cos α, and cos α.

Hence, the direction cosines of the vector which are equally inclined to the axes are

Question 12:

Let

and

. Find a vector

which is perpendicular to both

and

, and

Answer:

Let

Since

is perpendicular to both

and

, we have:

Also, it is given that:

On solving (i), (ii), and (iii), we get:

Hence, the required vector is

Question 13:

The scalar product of the vector

with a unit vector along the sum of vectors

and

is equal to one. Find the value of

Answer:

Therefore, unit vector along

is given as:

Scalar product of

with this unit vector is 1.

Hence, the value of λ is 1.

Question 14:

are mutually perpendicular vectors of equal magnitudes, show that the vector

is equally inclined to

and

Answer:

Since

are mutually perpendicular vectors, we have

It is given that:

Let vector

be inclined to

at angles

respectively.

Then, we have:

Now, as

Hence, the vector

is equally inclined to

Page No 459:

Question 15:

Prove that

, if and only if are perpendicular, given

Answer:

Question 16:

If θ is the angle between two vectors

and

, then only when

(A)

(B)

(C)

(D)

Answer:

Let θ be the angle between two vectors

and

Then, without loss of generality,

and

are non-zero vectors so that

It is known that

Hence, when

The correct answer is B.

Question 17:

Let

and

be two unit vectors andθ is the angle between them. Then

is a unit vector if

(A)

(B)

(C)

(D)

Answer:

Let

and

be two unit vectors andθ be the angle between them.

Then,

Now,

is a unit vector if

Hence,

is a unit vector if

The correct answer is D.

Question 18:

The value of

(A) 0 (B) –1 (C) 1 (D) 3

Answer:

The correct answer is C.

Question 19:

If θ is the angle between any two vectors

and

, then

when θisequal to

(A) 0 (B)

(C)

(D) π

Answer:

Let θ be the angle between two vectors

and

Then, without loss of generality,

and

are non-zero vectors, so that

Hence,

when θisequal to

The correct answer is B.

Breaking

NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra Miscellaneous Exercise